# One Way ANOVA Table for N = 18, k = 3 & α = 5% One way ANOVA table or summary to estimate the test of hypothesis (H0) at 5% or 0.05 significance level for sample means by using F-distribution table

One-Way ANOVA Classification Table
Sources of VariationdfSSMSSF-Ratio
between Treatments210.411110.4111/2

= 5.2056
5.2056/3.2467

= 1.6034
Error1548.748.7/15

= 3.2467

Total17

## Work with Steps for N = 18, k = 3 & α = 5%

The below is the example work with steps shows how to generate the one way ANOVA table for the different means of treatments to find the F test statistic to test hypothesis (H0) at a 5% or 0.05 significance level for three sample means x1, x2 & x3.

Workout :

step 1 Address the formula input parameters and values
x1 = 10, 12, 13, 11, 10, 13 & 15
x2 = 9, 11, 13, 9 & 11
x3 = 11, 15, 13, 12, 14 & 10

step 2 Form the below table to carry out the analysis of variance

 Total square x1 10 12 13 11 10 13 15 ∑x1 = 84 (∑x1)2 = 7056 x2 9 11 13 9 11 ∑x2 = 53 (∑x2)2 = 2809 x3 11 15 13 12 14 10 ∑x3 = 75 (∑x3)2 = 5625 G = 212 ∑Ti2 = 15490

Individual Squares :
 Total x12 100 144 169 121 100 169 225 ∑x12 = 1028 x22 81 121 169 81 121 ∑x22 = 573 x32 121 225 169 144 196 100 ∑x32 = 955 ∑xij2 = 2556

step 3 Find Correction Factor (CF)
N = total number of elements in all three sample dataset.
N = 7 + 5 + 6
N = 18
Correction factor (CF) = G2N
CF = 212218
= 44944/18

CF = 2496.8889

step 4 Find the sum of squares of all individual items (∑xij) and then total sum of squares (TSS)
Total sum of squares (TSS) = ∑∑xij2 - CF
= 2556 - 2496.8889
TSS = 59.1111

step 5 Find the sum of squares between the treatments (SST)

SST = k i = 1 Ti2ni - CF
= (∑x1)2n1+(∑x2)2n2+(∑x3)2n3 - CF
= 7056/7+2809/5+5625/6 - 2496.8889
= (1008 + 561.8 + 937.5) - 2496.8889
= 2507.3 - 2496.8889
= 10.4111

step 6 Find sum of squares due to the error (SSE)
SSE = TSS - SST
= 59.1111 - 10.4111
= 48.7

step 7 The above sum of squares together with their respective degrees of freedom and mean sum of squares will be summarized in the following table.

 Sources of Variation df SS MSS F-ratio between Treatments k - 1 SST SST/k-1 = MST MST/MSE = F T Error N - k SSE SSE/N-k = MSE Total N - 1

One-Way ANOVA Classification Table
Sources of Variation   df   SSMSSF-Ratio
between Treatment 3-1 = 210.411110.4111/2

= 5.2056
5.2056/3.2467

= 1.6034
Error 18-3 = 1548.748.7/15

= 3.2467

Total 18-1 = 17

step 8 Find Critical value of Fe or Table value of F
The critical value of Fe from the F-distribution table for the degrees of freedom (k - 1, N - k) at 5% significance level.
For treatments, the critical value Fe(2,15) from F-distribution table at 5% of significance level is 3.6823

Inference
There is no significance difference between treatments, since the calculated value of F0 = 1.6034 is smaller than the table value of Fe = 3.6823. Therefore the null hypothesis H0 is accepted. 